Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<set>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gt getchar()
#define ll long long
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
inline int in()
{
	int k=0;char ch=gt;bool p=1;
	while(ch<'-')ch=gt;if(ch=='-')ch=gt,p=0;
	while(ch>'-')k=k*10+ch-'0',ch=gt;
	return p?k:-k;
}
const int N=3e5+5;
int lz[N<<2];
#define lc k<<1
#define rc k<<1|1
struct node
{
	int l,r,cov;
	node(int l=0,int r=0,int cov=0):l(l),r(r),cov(cov){}
	inline bool operator<(const node &t)const{return l<t.l;}
};
typedef std::set<node>::iterator P;
std::set<node>cx;
P split(int l)
{
	P it=cx.lower_bound(l);
	if(it!=cx.end()&&it->l==l)return it;
	--it;node now=*it;cx.erase(it);
	cx.insert(node(now.l,l-1,now.cov));
	return cx.insert(node(l,now.r,now.cov)).first;
}
inline void cov(int l,int r,int x)
{	
	if(l>r)return;P itr=split(r+1),itl=split(l);
	cx.erase(itl,itr);cx.insert(node(l,r,x));
}
ll a[N],b[N],s[N],f[N],K;
int tot,L[N],R[N],ls[N];
inline int MO(const int &x){return x>=K?x-K:x;}
inline ll query(int p)
{
	P itl=split(p);int x=itl->cov;
	return (x?f[x]+MO(L[x]-p+K):0);
}
int main()
{
	int n=in();K=in();
	for(int i=1;i<=n;++i)
	{
		s[i]=s[i-1]+(a[i]=in()),b[i]=in();
		if(b[i]==2)L[i]=0,R[i]=K-1;
		else L[i]=MO(K-s[i-1]*2%K),R[i]=MO(K-s[i]*2%K);
		if(b[i]==1&&a[i]*2>K)return puts("-1"),0;
	}
	cx.insert(node(0,K-1,0));
	for(int i=n;i>=1;--i)
	{
		f[i]=query(L[i]);
		if(L[i]>R[i])cov(R[i]+1,L[i]-1,i);
		else cov(0,L[i]-1,i),cov(R[i]+1,K-1,i);
	}
	ll ans=f[1];
	for(int i=1;i<=n;++i)ans=std::min(ans,query(L[i]));
	for(int i=1;i<=n;++i)ans=std::min(ans,query(R[i]));
	printf("%lld\n",ans+s[n]*2);
	return 0;
}